lattice pie - another puzzle stuck on

is it possible to have an equilateral triangle in the plane all of whose vertices lie on the integer lattice{(m,n)|m,n are integers}? If so, what is the shortest side lengh possible for sucha triangle? what about other regular polygons?

Re: lattice pie - another puzzle stuck on

Originally Posted by sonia1

is it possible to have an equilateral triangle in the plane all of whose vertices lie on the integer lattice{(m,n)|m,n are integers}? If so, what is the shortest side lengh possible for sucha triangle? what about other regular polygons?

The answer is no. There is no such triangle exists.
It easy to prove that fact knowing that the area of regular triangle with side is that is irrational.

For , is a square which we can describe on system of lattice points.

For we can not describe the regular polygon on system of lattice points.

Here is a proof for this fact:

Suppose that there is regular polygon with n sides, that his edges are on lattice points. Now, from all regular polygons that can be describe on system of lattice points we choose the one with the minimal side length.

Say that the chosen one is: .

Now, we look at the reflections of for vis-a-vis (where and ).

So, we got a regular polygon with sides .

for is lattice point, why?

Due to our contraction is a rhombus.

Now, let and .

Let say that is intersection of the diagonals in rhombus.

We calculate that with two ways,

First with :

Ans with :

Hence,

We deduce from the above that and are integers.

Finally, he have regular polygon with sides witch his edges on lattice points.

But, (By our constraction) and this is a contradiction to the fact that chosen with minimal .

Re: lattice pie - another puzzle stuck on

In fact, it's impossible for an equilateral triangle to have all three vertices lying on . There are a few high-tech proofs (most notably a counting argument) but it can be done with simple geometry

Spoiler:

Without loss of generality assume that one of the vertices of our triangle is the orgin . Thus we are really examining , where are the other verticies. We may assume that for some . From here we break this into two cases, namely: , or . Clearly both cannot be zero since the origin and are distinct points.

Case 1, : It is clear that we assume WLOG that and for all other cases are analgous. Thus this triangle will have and for some . Clearly seeing though that the length of is (since they are both on the y-axis) we may conclude that the length of is as well. Realizing though that the lengths of are given independently by the distance equations and appealing once again to the equilaterallness of the triangle we may conclude that, in fact, . Solving this equation yields . As stated earlier though and focusing mainly on we see that which is clearly irrational for any rational .

Case 2, : Now we have that where neither or is zero. Constructing the midpoint of gives us that and that the length of the segment is . Now, form to be the median connecting and . If we consider the two segments to be the restrictions of the lines respectively we can see that and since we may conclue that (now you see why we needed ). Letting for some again and realizing that both lie on we may conclude that where represents the change in between the points and . Appealing to this we see that . Remembering though that forming the median bisected we see that the new triangle formed, is a triangle. In particular, this tells us that . Equating our two lengths of gives us and since we may conclude that . Thus, substitution into gives us that which is clearly irrational since .